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Consider the following matrix.$$\begin{pmatrix} 0 & 1 & 1 & 1\\ 1& 0& 1 & 1\\ 1& 1 & 0 & 1 \\1 & 1 & 1 & 0 \end{pmatrix}$$The largest eigenvalue of the above matrix is __________.

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$3$ is the correct answer.

Let $A = \begin{pmatrix} 0 & 1 & 1 & 1\\ 1& 0& 1 & 1\\ 1& 1 & 0 & 1 \\1 & 1 & 1 & 0 \end{pmatrix}$

Characteristic equation $\mid A – \lambda I \mid = 0$

$\implies \begin{vmatrix} -\lambda & 1 & 1 & 1\\ 1& -\lambda & 1 & 1\\ 1& 1 & -\lambda & 1 \\1 & 1 & 1 & -\lambda \end{vmatrix} = 0$

Perform the operation, $C_{4} \rightarrow C_{1} + C_{2} + C_{3} + C_{4}$, we get

$\implies \begin{vmatrix} -\lambda & 1 & 1 & 3-\lambda\\ 1& -\lambda & 1 & 3-\lambda\\ 1& 1 & -\lambda & 3-\lambda \\1 & 1 & 1 & 3-\lambda \end{vmatrix} = 0$

$\implies (3 – \lambda) \begin{vmatrix} -\lambda & 1 & 1 & 1 \\ 1& -\lambda & 1 & 1\\ 1& 1 & -\lambda & 1 \\1 & 1 & 1 & 1 \end{vmatrix} = 0$

Perform the operation, $R_{1} \rightarrow R_{1} - R_{2} , R_{2} \rightarrow R_{2} - R_{1}, R_{3} \rightarrow R_{3} - R_{1},$  we get

$\implies (3 – \lambda) \begin{vmatrix} -\lambda - 1 & 0 & 0 & 0 \\ 1 + \lambda & -\lambda-1 & 0 & 0 \\ 1 + \lambda & 0 & -\lambda – 1 & 0 \\1 & 1 & 1 & 1 \end{vmatrix} = 0$

$\implies (3-\lambda) (-\lambda – 1)(-\lambda – 1)(-\lambda – 1) = 0$

$\implies \lambda = -1,-1,-1,3$

$\therefore$ The largest eigenvalue is $3.$

$\textbf{PS:}$  For any matrix $A,$

• The determinant of $A$ equals the product of its eigenvalues.
• The trace of $A$ equals the sum of its eigenvalues.
• The trace of a matrix is defined as the sum of the leading diagonal entries.
• A real symmetric matrix has only real eigenvalues.

$$\text(OR)$$

Let $A = \begin{pmatrix} 0 & 1 & 1 & 1\\ 1& 0& 1 & 1\\ 1& 1 & 0 & 1 \\1 & 1 & 1 & 0 \end{pmatrix}$

We can write given matrix as $:3\begin{pmatrix} 0 &\frac{1}{3} &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &0 &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &0 &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &\frac{1}{3} &0 \end{pmatrix} = 3A$

A Markov matrix is a square matrix with all nonnegative entries, and where the sum of the entries down any column is $1.$ If the entries are all positive, it’s a positive Markov matrix. It is also called a doubly stochastic matrix.

The most important facts about a positive Markov matrix are:

• $\lambda = 1$ is an eigenvalue.
• The eigenvector associated with $\lambda = 1$ can be chosen to be strictly positive.
• All other eigenvalues have a magnitude less than $1.$

So, the correct answer is $3.$

References:

Thanks Lakshman.Is there any short method for finding the largest eigenvalue?

If you understand the structure of the matrix, you can solve it within some seconds.

You can write given matrix as : $3\begin{pmatrix} 0 &\frac{1}{3} &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &0 &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &0 &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &\frac{1}{3} &0 \end{pmatrix} = 3A$

This special matrix A is called Doubly Stochastic Matrix and it has largest eigen value $= 1$.

It is also used in Google’s PageRank algorithm.

Can i convert  the given matrix into upper triangular matrix(using Row Echileon) and see the diagonal element to find the largest element
Yes, you can. Show the process here.

eigen values is 3, -1,-1,-1 and maximun is 3